# Thread: 2018 HSC Marathon 3U Q and A

1. ## 2018 HSC Marathon 3U Q and A

I thought it would be nice to have a thread for answering past HSC questions and any other for that matter, in the lead up to the exam in a few weeks. So here it is.

I'll start off with one from the 2001 HSC which I have been struggling with. Any help would be appreciated.

http://prntscr.com/kumrq3

2. ## Re: 2018 HSC Marathon 3U Q and A

yea same having trouble with the second part

3. ## Re: 2018 HSC Marathon 3U Q and A

Consider $\triangle ATP$.

$\sin \alpha = \frac{h}{TP}$

$\sin \frac{\pi}{4} = \frac{h}{AT}= \frac{1}{\sqrt2}$

By the cosine rule,

$AP^2 = 2h^2 + h^2 {\rm cosec}^2\, \alpha - 2\sqrt2\, h^2\, {\rm cosec\,} \alpha \cos\theta$

Equating and dividing out by $h^2$ (the height cannot be zero),

\begin{aligned} \color{red}1 + \cot^2\alpha\color{black} - \cot \alpha &= 2 +\color{red}{\rm cosec}^2\, \alpha\color{black} - 2\sqrt2\, {\rm cosec\,} \alpha \cos\theta \\ -\cot \alpha &= 2-2\sqrt2\, {\rm cosec\,} \alpha \cos\theta \\ \cot \alpha &= 2\sqrt2\, {\rm cosec\,} \alpha \cos\theta -2 \\ \cos \alpha &= 2 \sqrt 2 \cos \theta - 2 \sin \alpha \\ \cos \theta &= \frac{1}{\sqrt 2}\sin \alpha + \frac{1}{2\sqrt 2} \cos \alpha \end{aligned}

4. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by HoldingOn
I thought it would be nice to have a thread for answering past HSC questions and any other for that matter, in the lead up to the exam in a few weeks. So here it is.

I'll start off with one from the 2001 HSC which I have been struggling with. Any help would be appreciated.

http://prntscr.com/kumrq3
Could someone try part iii as well

5. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by HoldingOn
Could someone try part iii as well
$\cos \theta =\frac{1}{2\sqrt 2} \cos \alpha + \frac{1}{\sqrt 2}\sin \alpha$

Applying the auxiliary angle transformation:

\begin{aligned} \cos \theta &= \sqrt{\left( \frac{1}{\sqrt 2}\right)^2 + \left( \frac{1}{2\sqrt 2} \right)^2} \cos \left(\alpha - \tan ^{-1} \frac {\left( \frac{1}{\sqrt2} \right) }{\frac 12\left(\frac{1}{\sqrt2}\right) }\right) \\ &= \sqrt{\frac58} \cos \left(\alpha - \tan ^{-1} 2\right) \\ \theta &=\cos^{-1}\left(\sqrt{\frac58} \cos \left(\alpha - \tan ^{-1}2\right)\right)\quad (*)\end{aligned}

* Note the restriction $0 < \theta < \pi$ in the diagram - we do not need to take a general solution here.

$\alpha = 0, \quad \theta \approx 1.21$

$\alpha = \frac \pi 2, \quad \theta = \frac \pi 4$.

$\frac{d\theta}{d\alpha} = \left(-\frac{1}{\sqrt{1 - \cos^2 \theta}}\right)\left(-\sqrt \frac 58 \sin\left(\alpha - \tan^{-1} 2\right)\right)$

$\frac{d\theta}{d\alpha} = 0, \quad \alpha = \tan^{-1} 2 \approx 1.11, \quad \theta \approx 0.66$

(Since $0 < \alpha < {\pi}/{2}$)

Since this is the only turning point within the given domain and $0.66 < \pi/4, \,\,0.66 < 1.21$, this is clearly a minimum turning point.

If we put $\theta$ on the vertical axis, the graph of this function will be similar in shape to a concave up parabola.

Note when sketching that $0,\,\,\pi/2$ are actually excluded from the domain, so you should draw an open circle at the endpoints of the domain to indicate that they are not part of the function.

7. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by HoldingOn
i) you expand $x(1+x)^n$ , differentiate both sides then sub in x = 1

ii) you expand $(1+x)^n$, integrate both sides twice (make sure you can find the constants) then sub x = -1 and do some re arrangement.

8. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by integral95
i) you expand $x(1+x)^n$ , differentiate both sides then sub in x = 1

ii) you expand $(1+x)^n$, integrate both sides twice (make sure you can find the constants) then sub x = -1
Ok thanks I'll try that. Just out of curiosity is there anything you look for when approaching these more left-field binomials, or just intuition?

9. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by HoldingOn
Ok thanks I'll try that. Just out of curiosity is there anything you look for when approaching these more left-field binomials, or just intuition?
Mostly the latter for me really, there's a lot of pattern recognition going on when I approach them.

10. ## Re: 2018 HSC Marathon 3U Q and A

http://prntscr.com/kw5d3q

Part iii. Thanks

11. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by HoldingOn
http://prntscr.com/kw5d3q

Part iii. Thanks
The horizontal range of the ball is maximised when $\alpha = \pi / 4$. This can be easily shown by examining the equation in ii).

Case 1: The ceiling allows for an angle of projection of $\pi / 4$.

If the ceiling allows for at least this angle of projection, then $\alpha = \pi / 4$ is the optimal angle to throw the ball.

It doesn't matter how tall the ceiling is after a certain point - because the best angle is $\pi / 4$ and any higher angle will simply lower $d$.

The horizontal range obtained by projecting at this optimal angle is then $d = {v^2}/{g}$.

But when does the ceiling allow for this angle? It is when the maximum height of the projectile (with angle of projection $\alpha = \pi / 4$) is less than or equal to the height allowed by the ceiling (which is $H-S$).

Mathematically, this can be represented by the condition

$(H-S) \geq \frac{v^2 \sin^2 (\pi/4)}{2g} \implies 4g(H-S) \geq v^2$.

Case 2: The ceiling does NOT allow for an angle of projection of $\pi / 4$.

As $\alpha$ decreases past $\pi / 4$, so too does $d$, from ii).

Therefore we want the largest value of $\alpha$ we can get. This value occurs when the tip of the ball's trajectory coincides with the ceiling, i.e.

$(H-S) = \frac{v^2 \sin^2 \alpha}{2g}$

Rearrangement gives:

$\sin^2 \alpha = \frac{2g(H-S)}{v^2} \implies \sin \alpha = \sqrt{\frac{2g(H-S)}{v^2}} \quad (1)$

$\cos^2 \alpha = 1 - \frac{2g(H-S)}{v^2} \implies \cos \alpha = \sqrt{\frac{v^2 - 2g(H-S)}{v^2}} \quad (2)$

Finally, to get the answer, substitute (1) and (2) into:

$d = \frac{v^2}{g}(2 \sin \alpha \cos \alpha)$

12. ## Re: 2018 HSC Marathon 3U Q and A

Could I get some help with part i and ii please. Thanks

http://prntscr.com/kz7x4d

13. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by HoldingOn
Could I get some help with part i and ii please. Thanks

http://prntscr.com/kz7x4d
CCI27092018_00000.jpg

14. ## Re: 2018 HSC Marathon 3U Q and A

i) Hint: the expression you need to prove can be rearranged to

$\cot \theta = \frac{1 + \tan n\theta\, \tan (n+1)\theta}{\tan(n+1)\theta - \tan n\theta}$

ii) Starting from the inductive hypothesis:

$\sum_{n= 1}^k \tan \theta\tan (n+1)\theta = -(k+1) + \cot \theta \tan (k+1)\theta$

$\sum_{n= 1}^{k+1} \tan \theta\tan (n+1)\theta = -(k+1) + \cot \theta \tan (k+1)\theta +\color{red} \tan(k+1)\theta \tan(k+2)\theta$

From i) by rearranging we get $\color{red} \tan n\theta\, \tan (n+1)\theta \color{black}= \cot \theta(\tan(n+1)\theta - \tan n\theta) - 1$.

15. ## Re: 2018 HSC Marathon 3U Q and A

This seems to be my thread haha.

If anyone could help with this, and maybe some tips on how to approach these sort of problems- I'm not great at them.

A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is four times the probability of getting 2 even numbers. Find the probability that a single throw of the die results in an even number.

16. ## Re: 2018 HSC Marathon 3U Q and A

This is a Binomial Probability question.

$\binom 8 3 p^3 (1-p)^5 = 4 \times \binom 8 2 p^2(1-p)^6$

Solving you get the P(even) in a single throw = p = 2/3.

17. ## Re: 2018 HSC Marathon 3U Q and A

May I ask where you found that proj motion question?

18. ## Re: 2018 HSC Marathon 3U Q and A

Originally Posted by herowise
may i ask where you found that proj motion question?
2003 3U paper

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•